Determine the mass of silver chloride precipitated by the interaction of 200 grams of hydrochloric acid solution

| January 8, 2021 | education

Determine the mass of silver chloride precipitated by the interaction of 200 grams of hydrochloric acid solution with a concentration of 24.6% with an excess of silver nitrate solution.

1. Let’s write the reaction equation:
HCl + AgNO3 = AgCl ↓ + HNO3;
2. find the mass of hydrogen chloride in the solution:
m (HCl) = w (HCl) * m (solution HCl);
m (HCl) = 0.246 * 200 = 49.2 g;
3.Calculate the chemical amount of HCl:
n (HCl) = m (HCl): M (HCl);
M (HCl) = 1 + 35.5 = 36.5 g / mol;
n (HCl) = 49.2: 36.5 = 1.348 mol;
4. from the recording of the reaction equation it is seen:
n (AgCl) = n (HCl) = 1.348 mol;
5.find the mass of the sediment:
m (AgCl) = n (AgCl) * M (AgCl);
M (AgCl) = 108 + 35.5 = 143.5 g / mol;
m (AgCl) = 1.348 * 143.5 = 193.44 g.
Answer: 193.44 g.



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